Re: [Haskell-cafe] manage effects in a DSL

3 Feb 2014

      Hi Corentin,

Oleg's January 31 message has: `ReadAccount  :: Exp r Int'. If you use that
version you can define that last `hasEffect :: Exp Effect ()'. Then bind
can just be:

Bind :: Exp m a -> (a -> Exp m b) -> Exp m b

And the `m' from a non-effectful thing (ReadAccount) will be set to Effect
when you bind it with an effectful computation. You can still have
operations that take arguments like `Exp NoEffect a', which will give you a
type error when you pass in a an argument tagged with Effect.

Adam

On Mon, Feb 3, 2014 at 6:44 AM, Corentin Dupont
wrote:
...
I saw that to write liftQD you decontruct (unwrap) the type and
reconstruct it.
I don't know if I can do that for my Exp (which is a full DSL)...
Anyway, there should be a way to encode the Effect/NoEffect semantic at
type level...
Using Oleg's parametrized monad idea (
http://hackage.haskell.org/package/monad-param-0.0.2/docs/Control-Monad-Para...),
I tried:
...
{-# LANGUAGE KindSignatures, DataKinds, ScopedTypeVariables, GADTs
  MultiParamTypeClasses, FunctionalDependencies, FlexibleInstances,
UndecidableInstances #-}
...
module DSLEffects where
import Prelude hiding (return, (>>), (>>=))
import Control.Monad.Parameterized
This data type will be promoted to kind level (thanks to DataKinds):
...
data Eff = Effect | NoEffect
This class allows to specify the semantic on Effects (Effect + NoEffect =
Effect):
...
class Effects (m :: Eff) (n::Eff) (r::Eff) | m n -> r
instance Effects Effect n Effect
instance Effects NoEffect n n
This is the DSL:
...
data Exp :: Eff -> * -> * where
  ReadAccount  :: Exp NoEffect Int      --ReadAccount has no effect
  WriteAccount :: Int -> Exp Effect ()  --WriteAccount has effect
  Const        :: a -> Exp r a
  Bind         :: Effects m n r => Exp m a -> (a -> Exp n b) -> Exp r b
--Bind comes with a semantic on effects
  Fmap         :: (a -> b) -> Exp m a -> Exp m b
...
instance Functor (Exp r) where
  fmap = Fmap
...
instance Return (Exp r) where
   returnM = Const
...
instance (Effects m n r) => Bind (Exp m) (Exp n) (Exp r) where
   (>>=) = Bind
...
noEff :: Exp NoEffect ()
noEff = returnM ()
...
hasEffect :: Exp Effect ()
hasEffect = ReadAccount >> (returnM () :: Exp Effect ())
This is working more or less, however I am obliged to put the type
signature on the returnM (last line): why?
Furthermore, I cannot write directly:
...
hasEffect :: Exp Effect ()
hasEffect = ReadAccount
Do you have a better idea?
On Sun, Feb 2, 2014 at 8:55 PM, Lindsey Kuper wrote:
...
On Sun, Feb 2, 2014 at 2:42 PM, Corentin Dupont
 wrote:
...
you should be able to run an effectless monad in an effectful one.
How to encode this semantic?
In LVish we just have a `liftQD` operation that will let you lift a
deterministic computation to a quasi-deterministic one (recall that
deterministic computations can perform fewer effects):
liftQD :: Par Det s a -> Par QuasiDet s a
So, analogously, you could have a `liftEff` and then write `liftEff
noEff`.  This is also a little bit ugly, but you may find you don't
have to do it very often (we rarely use `liftQD`).
Lindsey
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Re: [Haskell-cafe] manage effects in a DSL

adam vogt